calculate-I-0-pi-2-4-tanx-log-tanx-dx-and-J-0-pi-2-log-tanx-3-tanx-dx-

Question Number 141218 by mathmax by abdo last updated on 16/May/21

calculate I = \int_{0}^{\frac{π}{2}}^{4} \sqrt{tanx} \log (tanx) dx and

J = \int_{0}^{\frac{π}{2}} \frac{\log (tanx)}{(^{3} \sqrt{tanx})} dx

Answered by Ar Brandon last updated on 16/May/21

f (α) = \int_{0}^{\frac{π}{2}} \tan^{α} xdx = \frac{1}{2} Γ (\frac{α + 1}{2}) Γ (\frac{1 - α}{2}) = \frac{π}{2 \sin (\frac{1 - α}{2} π)}

f (\frac{1}{4}) = \frac{Γ (\frac{5}{8}) Γ (\frac{3}{8})}{2} = \frac{π}{2 \sin (\frac{3 π}{8})} ★

\ln (f (α)) = \ln (\frac{π}{2}) - lnsin (\frac{1 - α}{2} π)

\frac{f^{'} (α)}{f (α)} = \frac{π}{2} \cdot \tan (\frac{1 - α}{2} π) \Rightarrow f^{'} (α) = f (α) {\frac{π}{2} \cdot \tan (\frac{1 - α}{2} π)}

f^{'} (\frac{1}{4}) = f (\frac{1}{4}) {\frac{π}{2} \cdot \tan (\frac{3 π}{8})}

I = \frac{π^{2}}{4} \tan (\frac{3 π}{8}) cosec (\frac{3 π}{8})

Answered by Ar Brandon last updated on 16/May/21

f (α) = \int_{0}^{\frac{π}{2}} \tan^{α} xdx = \frac{1}{2} Γ (\frac{α + 1}{2}) Γ (\frac{1 - α}{2}) = \frac{π}{2 \sin (\frac{1 + α}{2} π)}

f (- \frac{1}{3}) = \frac{π}{2 \sin (\frac{π}{3})} ★

\ln (f (α)) = \ln (\frac{π}{2}) - lnsin (\frac{1 + α}{2} π)

\frac{f^{'} (α)}{f (α)} = \frac{π}{2} \cdot \tan (\frac{1 + α}{2} π) \Rightarrow f^{'} (α) = f (α) {\frac{π}{2} \cdot \tan (\frac{1 + α}{2} π)}

f^{'} (- \frac{1}{3}) = f (- \frac{1}{3}) {\frac{π}{2} \cdot \tan (\frac{π}{3})}

J = \frac{π^{2}}{4} \tan (\frac{π}{3}) cosec (\frac{π}{3})

Commented by Ar Brandon last updated on 16/May/21

f^{'} (α) = \frac{π^{2}}{4} \tan (\frac{1 - α}{2} π) cosec (\frac{1 - α}{2} π)

Answered by mathmax by abdo last updated on 16/May/21

let f (a) = \int_{0}^{\frac{π}{2}} {(tanx)}^{a} dx \Rightarrow f (a) = \int_{0}^{\frac{π}{2}} e^{aln (tanx)} dx \Rightarrow

f^{^{'}} (a) = \int_{0}^{\frac{π}{2}} {(tanx)}^{a} \log (tanx) dx \Rightarrow f^{^{'}} (\frac{1}{4}) = \int_{0}^{\frac{π}{2}} {(tanx)}^{\frac{1}{4}} \log (tanx) dx = I

changement tanx = t give f (a) = \int_{0}^{\infty} \frac{t^{a}}{1 + t^{2}} dt =_{t = \sqrt{z}} \int_{0}^{\infty} \frac{z^{\frac{a}{2}}}{1 + z} \frac{dz}{2 \sqrt{z}}

= \frac{1}{2} \int_{0}^{\infty} \frac{z^{\frac{a}{2} - \frac{1}{2}}}{1 + z} dz = \frac{1}{2} \int_{0}^{\infty} \frac{z^{\frac{a + 1}{2} - 1}}{1 + z} dz = \frac{π}{2 \sin (π (\frac{a + 1)}{2}))}

= \frac{π}{2 \cos (\frac{π a}{2})} cv of f (a) we must have 0 < \frac{a + 1}{2} < 1 \Rightarrow - 1 < a < 1

f^{^{'}} (a) = - \frac{π}{2} \times \frac{- \frac{π}{2} \sin (\frac{π a}{2})}{\cos^{2} (\frac{π a}{2})} = \frac{π^{2}}{4} . \frac{\sin (\frac{π a}{2})}{\cos^{2} (\frac{π a}{2})}

I = f^{^{'}} (\frac{1}{4}) = \frac{π^{2}}{4} . \frac{\sin (\frac{π}{8})}{\cos^{2} (\frac{π}{8})} = \frac{π^{2}}{4} . \frac{\frac{\sqrt{2 - \sqrt{2}}}{2}}{\frac{2 + \sqrt{2}}{4}} = \frac{π^{2}}{2} \times \frac{\sqrt{2 - \sqrt{2}}}{2 + \sqrt{2}}

J = \int_{0}^{\frac{π}{2}} {(tanx)}^{- \frac{1}{3}} \log (tanx) dx = f^{^{'}} (- \frac{1}{3})

= - \frac{π^{2}}{4} . \frac{\sin (\frac{π}{6})}{\cos^{2} (\frac{π}{6})} = - \frac{π^{2}}{4} \times \frac{\frac{1}{2}}{{(\frac{\sqrt{3}}{2})}^{2}} = - \frac{π^{2}}{8} \times \frac{4}{3}

= - \frac{π^{2}}{6}